- Let $a(n)$ be A217061. Here $$ a(n) = \sum\limits_{m=1}^{n}\frac{1}{(m-1)!}\sum\limits_{k=0}^{n-m}(n+k-1)!\sum\limits_{j=0}^{k}\frac{1}{(k-j)!}\sum\limits_{\ell=0}^{j}\frac{2^{\ell-j}(-1)^{\ell+j}s(n-m-\ell+j, j-\ell)}{\ell!(n-m-\ell+j)!}. $$ where $s(n,k)$ are (signed) Stirling numbers of the first kind.
- Start with vector $\nu$ of fixed length $m$ with elements $\nu_i=1$ (that is, $\nu=\{1,1,\dotsc,1\}$), reserve $t$ as an empty vector of fixed length $m$, set $t:=\nu$ and for $i$ from $1$ to $m-1$, for $j$ from $1$ to $m-i$ consecutively apply $$ \nu_{j+1} := j\nu_j + \nu_{j+1}, \\ \nu_{j} := j\nu_j + \nu_{j+1}. $$ We also need to apply $t_{i+1} = \nu_1$ (after ending each cycle for $j$).
I conjecture that after the whole transform $$ t_n = a(n). $$
Here is the PARI/GP program to check it numerically:
a(n) = sum(m=1, n, 1/(m-1)!*sum(k=0, n-m, (n+k-1)!*sum(j=0, k, 1/(k-j)!*sum(l=0, j, 2^(j-l)*(-1)^(l+j)*stirling(n-m-l+j, j-l, 1)/(l!*(n-m-l+j)!)))))
upto1(n) = my(v1); v1 = vector(n, i, 1); v2 = v1; for(i=1, n-1, for(j=1, n-i, v1[j+1] += j*v1[j]; v1[j] = j*v1[j] + v1[j+1]); v2[i+1] = v1[1]); v2
test1(n) = upto1(n) == vector(n, i, a(i))
In addition, this question can be rephrased as follows:
- Let $$ R(n, q) = \begin{cases} 1 & \textrm{if } n = 0 \\ R(n-1, q+1) + 2(q+1)!\sum\limits_{j=0}^{q} \frac{R(n-1,j)}{j!} & \textrm{otherwise} \end{cases} $$
I conjecture that $$ R(n,0)=a(n+1). $$
Is there a way to prove it?
